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Consider a strictly increasing sequence $1\leq q_0<q_n<q_{n+1}<q$ such that $q_n\to q$ as $n\to \infty$. Let $B\subset \Bbb R^d$ be a ball, so that $L^{q}(B)\subset L^{q_{n+1}}(B)\subset L^{q_{n}}(B)\subset L^{q_0}(B)$. Consider a sequence $(u_n)_n\subset L^q(B)$ such that $\|u_n\|_{L^{q_n}(B)}=1$ and $u_n\to u$ in $L^{q_0}(B)$.

I want to prove or disprove that $\|u\|_{L^{q}(B)}=1$. Note that by Fatou's lemma, we get $$\|u\|_{L^{q}(B)}\leq \liminf_{n\to\infty}\|u_n\|_{L^{q_n}(B)}=1.$$

Addendum: I forgot the assumption that $d>q.$

The answer by @FedorPetrov gives a nice counter example when q>d.

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It is not true even if all $q_n$ are equal to $q$. Take $q=2$, $q_0=1$, $d=1$, $B=[0,1]$, $u_n=n^{1/2}\chi_{[0,1/n]}$, $u=0$.

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  • $\begingroup$ I acknowledge this counter example works. I forgot to mention that d>q.I encountered this problem in the context of Sobolev embedding. Sorry for the inconvenience. $\endgroup$
    – Guy Fsone
    Commented Jun 22, 2023 at 13:28
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    $\begingroup$ @GuyFsone : Essentially the same example, with $u(x_1,x_2,\dots,x_d)=\sqrt n\,1(0<x_1<1/n,0<x_2<1,\dots,0<x_d<1)$, works for any $d$. $\endgroup$
    – Iosif Pinelis
    Commented Jun 22, 2023 at 13:49

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